Question: You have found the following ages (in years) of all 6 bears at your local zoo: $ 18,\enspace 2,\enspace 2,\enspace 23,\enspace 21,\enspace 2$ What is the average age of the bears at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Because we have data for all 6 bears at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{18 + 2 + 2 + 23 + 21 + 2}{{6}} = {11.3\text{ years old}} $ Find the squared deviations from the mean for each bear. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $18$ years $6.7$ years $44.89$ years $^2$ $2$ years $-9.3$ years $86.49$ years $^2$ $2$ years $-9.3$ years $86.49$ years $^2$ $23$ years $11.7$ years $136.89$ years $^2$ $21$ years $9.7$ years $94.09$ years $^2$ $2$ years $-9.3$ years $86.49$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{44.89} + {86.49} + {86.49} + {136.89} + {94.09} + {86.49}} {{6}} $ $ {\sigma^2} = \dfrac{{535.34}}{{6}} = {89.22\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{89.22\text{ years}^2}} = {9.4\text{ years}} $ The average bear at the zoo is 11.3 years old. There is a standard deviation of 9.4 years.